Question: Let $f(x) = -10x^{2}-7x+8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-10x^{2}-7x+8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -10, b = -7, c = 8$ $ x = \dfrac{+ 7 \pm \sqrt{(-7)^{2} - 4 \cdot -10 \cdot 8}}{2 \cdot -10}$ $ x = \dfrac{7 \pm \sqrt{369}}{-20}$ $ x = \dfrac{7 \pm 3\sqrt{41}}{-20}$ $x =\dfrac{7 \pm 3\sqrt{41}}{-20}$